Free s-Block and p-Block Elements MCQs with Answers
1,359 s-Block and p-Block Elements MCQs from Chemistry, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
The s and p blocks are studied through their valence-shell configurations, periodic trends and characteristic chemical properties. Work covers the reactions of Group I and Group II elements, the behaviour of Group IV elements, and how atomic size, ionization energy, electronegativity and metallic character change across periods and down groups.
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1,359 questions · page 28 of 68
- A. Reducing power in aqueous solution is maximum for Lithium metal.
- B. electron affinity order O+<O<O-22<O-2
- C. pH of aqueous solution LiCl>BeCl2>MgCl2>AlCl3
Explanation: pH of aqueous solution LiCl<BeCl2<MgCl2<AlCL3 Therefore, option C is correct.
Correct answer: pH of aqueous solution LiCl>BeCl2>MgCl2>AlCl3- A. d-block.
- B. f-block.
- C. s-block.
- D. p-block.
Explanation: This electronic configuration is of Cr. Cr is a d-block element. Therefore, option A is correct.
Correct answer: d-block.- A. 1.36, 1.40, 1.71
- B. 1.36, 1.71, 1.40
- C. 1.71, 1.40, 1.36
- D. 1.71, 1.36, 1.40
Explanation: N-3 > O-2 > F-1.71 1.40 1.37Å Hence, option C is correct.
Correct answer: 1.71, 1.40, 1.36- A. 3, 11, 19, 37
- B. 5, 13, 21, 39
- C. 7, 15, 31, 49
- D. 5, 13, 31, 49
Explanation: The elements of group III A are: ⇒ 5B, 13Al, 31Ga, 49ln, 81Tl Therefore option D is correct.
Correct answer: 5, 13, 31, 49- A. Ni<Pd<Pt
- B. Ti<Zr<Hf
- C. Ti <Zr <Pb
- D. All are correct.
Explanation: Due to lanthanide contraction size of Pd ≈ Pt and size of Zr ≈ Hf (size of 5d series elements is ≈ size of 4d series elements).
Correct answer: Ti<Zr<Hf- A. I+<I <I- ⇒ order of size.
- B. Se <S <Cl ⇒ order of ionisation energy.
- C. F<O<N ⇒ order of oxidizing property
- D. Na <Mg <Al ⇒ order of electronegativity.
Explanation: N<O<F is correct order of oxidizing property, as oxidizing property increases in a period. Therefore option C is correct.
Correct answer: F<O<N ⇒ order of oxidizing property- A. Greater metallic character
- B. Larger atomic size
- C. Strong reducing agent
- D. Less electropositive
Explanation: The factors upon which the ionization energy of an atom mainly depends are magnitude of nuclear charge, size of the atom, and the…
Correct answer: Less electropositive- A. Al
- B. Si
- C. P
- D. S
Explanation: Sulfur is located below phosphorus in the periodic table, in the same group (Group 16).
Correct answer: S- A. The energy released when an electron is added to an atom of the element
- B. The same as the electron affinity of the element
- C. Equal in magnitude but of opposite sign to the electron affinity of the element
- D. The energy required when electron is removed from outermost shell.
Explanation: The ionization energy of an element is the energy needed to remove an electron from an atom's outermost shell.
Correct answer: The energy required when electron is removed from outermost shell.- A. IA
- B. VIA
- C. IIA
- D. VIIA
Explanation: Group VIIA elements have positive first electron affinity, indicating they tend to gain electrons easily. This is why option D is correct.
Correct answer: VIIA- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation: Option C is correct. The modern periodic table actually consists of seven periods and eighteen groups.
Correct answer: Option C- A. O > N
- B. Be > B
- C. Al > Mg
- D. Ne > He
Explanation: The correct answer is Option D: Ne > He. This statement is incorrect because helium has a higher first ionization energy than neon.
Correct answer: Ne > He- A. F > Cl > Br > I
- B. Cl > F > Br > I
- C. F < Cl < Br < I
- D. Cl < F < Br < I
Explanation: The correct order of decreasing electron affinities for Group VII (halogens) is:Cl > F > Br > IHere's why:General Trend: Electron affinity…
Correct answer: Cl > F > Br > I- A. Ionization potential
- B. Electronegativity
- C. Electron affinity
- D. Lattice energy
- E. Potential energy
Explanation: When an electron is added to a neutral gaseous atom to convert it into a negative ion, the enthalpy change accompanying the process is…
Correct answer: Electron affinity- A. Li is less electropositive
- B. Li has low ionization potential
- C. Li+ can effectively polarize CO3-2 ion
- D. Li+ cannot effectively polarize CO3-2 ion
Explanation: The thermal instability of lithium carbonate (Li2CO3) compared to other alkali metal carbonates due to following reasons:Size and Charge…
Correct answer: Li+ can effectively polarize CO3-2 ion- A. Basic oxides
- B. Acidic oxides
- C. Amphoteric oxides
- D. Both B and C
Explanation: Alkali metals are basic. The oxides they form are also basic due to this.
Correct answer: Basic oxides- A. Li+
- B. Na+
- C. K+
- D. Cs+
Explanation: When considering the heat of hydration, the size and charge density of cations play a crucial role.
Correct answer: Cs+- A. Positive
- B. Negative
- C. Zero
- D. All
Explanation: The heat of solution refers to the heat change when a solute dissolves in a solvent.
Correct answer: Negative- A. Low electronegativity
- B. High charge density
- C. Low charge density
- D. Not known yet
Explanation: Lithium ion (Li⁺) is very small with a high charge density. This high charge density polarizes the carbonate ion (CO₃²⁻) strongly, making…
Correct answer: High charge density- A. Sr(OH)2
- B. Mg(OH)2
- C. Ba(OH)2
- D. Ca(OH)2
Explanation: The solubility of alkaline earth metal hydroxides in water increases down the group. Be(OH)2 is quite insoluble.
Correct answer: Ca(OH)2